Drilling and Completions PETR8502: Assignment 1

Peter Smith 20724513

Lecturer:

Dr Khalil Rahman

Date Submitted:

15 October 2011 (via Dr Jian Guo Wang)

Question 1

1. Describe the merits and drawbacks of jackup rigs in the following contexts compared to

drillships [ 6 MARKS]:

a. Towing and positioning to the location

b. Stability/failure of the rig (both structural and foundation)

c. Moving to the next locations

d. Deck accommodation

e. Cost

f. Metocean conditions (current, cyclone, etc.)

Towing and Positioning to the location:

Drillships are self powered. Jackups need a secondary vessel (i.e. a tug boat, sometimes two) to

position on location. Drillships are designed with lateral thrusters (in newer ones, older ones use 8

point anchors and tensioners) that can aid positioning to the site.

Once on location however a Jackup will remain on site without difficulty, however a Drillship either

needs to consume energy to constantly maintain position (via Dynamic Positioning and thrusters or

via re-tensioning anchors). Drillships are prone to movement in heavy sea states whereas a Jackup

can be positioned high enough above the water to be unaffected greatly by stormy seas.

Stability / Failure of the Rig:

Drillships are inherently less stable as they are not directly attached (by gravity) to the sea floor.

They are prone to sea swell, heavy sea states and Typhoons. The author of this assignment has been

on a drill ship (“Deep Sea Duchess”) that suffered a failure of the anchoring system during heavy

seas caused by a typhoon with a subsequent failure of the marine riser and the subsea BOP

detaching itself from the wellhead offshore Vietnam.

Jackups, will being less prone to failure due to sea states and weather, a prone to stability issues if a

blowout in the well occurs below the cemented casing and the sea bed shifts or becomes porous.

The Sedco 702 almost sunk completely offshore due to a similar failure.

Jackups are not floating when on location; hence hull failure is not an issue during operations unlike

a drillship. Drillships are prone to overturning if not correctly balanced (i.e. if too much drill pipe is

kept on deck they can turn “turtle” in high wind).

Moving to next location:

One of the significant advantages of Jackups is in drilling a multi well template or pattern and drill

over platforms. They are more suited to this as they have a larger deck area and a drilling template

can be drilled through for a larger number of wells (40 is possible on larger rigs). For a Drillship to

drill a new well typically it has to be moved (repositioned) which required lifting the anchors,

repositioning and re-tensioning the anchors. However to leave the location and move to a new

location a drillship is far faster to mobilize as the prime mover is self contained. DPS drillships can be

moved almost instantaneously where as a Jackup needs to retract it’s legs, be towed by a tug and

then re-establish it’s stability.

Deck Accommodation

Jackups can be modified more easily and often have cantilevered sections added. A drill ship cannot

be extended easily. The nature of a drill ship is to have a narrow but long deck which limits

movement laterally whereas a Jackup has either a square or large rectangular work deck. A Jackup

can be positioned over an existing platform and drill through the platform while utilizing the

platform accommodation, this is not possible for a drillship.

However, because of their cargo-carrying capacity and exceptional mobility, drillships are especially

useful for drilling exploratory wells.

Cost

Jackups are generally cheaper than drill ships due to the larger supply and lower complexity.

However, regardless of cost drilling ships can work in deeper waters than Jackups and hence the two

rigs types are generally mutually exclusive to the market.

Metocean conditions

The long narrow hull of the drillship results in motion in all six degrees of freedom, particularly pitch,

roll and heave. Mooring systems and dynamic positioning systems have greatly improved the

drillship's capability of drilling in harsh environments, however drill ships must move off location

usually in typhoon and cyclone conditions.

There are two types of jackup rigs, the independent-leg type, usually three legs with lattice

construction and the mat type, in which the legs are attached to a very large mat that rests on the

ocean bottom. Having a mat on a muddy/ soft ocean floor might seem better option. than having

to drive the legs into oceans bottom (as with the independent leg), making the operation vulnerable

to weather changes. However, the mat is susceptible to damage by workboats, or to movement of

objects on the ocean bottom. These reasons have lead to a preference to the independent leg

jackup. Jackups however are not suitable for locations with strong constant currents.

2. Describe how the formation overpressure can be predicted while drilling by monitoring

various drilling parameters and discuss the uncertainties involved with such prediction. Also

discuss the d-exponent and modified d-exponent techniques to improve pore pressure prediction.

Formation overpressure is predicted prior to drilling by the use of seismic data analysis (which examines the interval transit time of sound waves which can predict pore pressure) Formation overpressure during drilling is predicted from; :

• Monitoring the drilling parameters (WOB, RPM, ROP, Torque, drag, overpull) If Weight on Bit and RPM (bit speed) are kept constant then an increase in less compacted rock. Less compacted rock may indicate a change in pore pressure caused by trapped gas. . However, WOB and RPM are rarely kept constant. So we calculate the ‘drillability’ of the formation, called the ‘d’ exponent assuming a constant mud weight. A sudden decreased in dexp indicates the potential existence of overpressure. Torque and drag can indicate a stuck bit or a bit entering a more porous zone. Overpull can indicate a change in cuttings caused by a change on pore pressure and the potential for a stuck bit.

• Mud and cuttings returning from the hole (mud logging) A change in cuttings type (to slivered cuttings) can indicate a breakthrough into a overpressured zone. A change in the rate of mud returns can also indicate potential overpressure. Higher returns indicate an increase in BHP.

• Electrical properties of the formation (electric logging) Changes in electrical properties indicate changes in conductivity (i.e. resistivity) caused by changes in salt (chloride concentrations). Chloride concentration changes in overpressured zones due to the increased hydrostatic pressure. Resistivity hence decreases where pore pressure increases. Additional logging data that can be used to detect overpressure include Sonic Velocity (rate of change becomes positive in overpressured zone), porosity (becomes gradually positive in an overpressured zone) and temperature (rate of increase sharpens in OP zone).

• Direct evidence (influx) Influx is the presence of overpressure pore fluid changing the rate of mud returns. in the presence of overpressure rock can slough into the hole and fail to be returned and affect the rate of return. Any change in the rate of return of mud should be taken as evidence of a change in the nature of the well.

• The d exponent and modified d exponent are mathematical techniques for determining pore pressure. The dexp formula is shown below (from the lecture notes).

In normal drilling dexp will increase with depth (as WOB increases and ROP decreases. Any sudden decrease in dexp may indicate a presence of overpressure. The modified dexp takes account of changing mud weight by multiplying by a normalized density (mud weight). The formula is below.

nρ = mud density for normal pressure and ecρ indicated the equivelant circulating density

at the bit.

• Uncertainties with these predictions.

A single parameter is not adequate for overpressure detection along. For example Mud return rates can change if other well failures occur, electrical logging may fail, ROP may change sharply if the bit fails or drill string collapses. WOB may change is mud density is changes unknowingly.

ec

nddρ

ρ×= expmod

3. Select appropriate bit types and discuss their geometric and operational features for the

following three different rock formations [3 + 2 MARKS]:

a. Very hard

b. Very soft

c. Medium hard.

Also discuss the consequences when inappropriate bit types are used.

a. A bit that is selected for a very hard formation would have an IADC code such as; 8-4-7 this is a roller bit designed with tungsten carbide teeth for a very hard formation with teeth designed for the hardest class of rock in the category fitted with friction, sealed bearing with gauge protection (for high WOB).

b. A bit for a very soft formation could have an IADC code such as; 4-1-1 This is a roller bit with tungsten carbide teeth for a soft formation on the lowest hardness of the first category and standard bearings. This bit would be used with low WOB and could be very large – i.e. 24” for drilling for a upper casing. An alternative lower cost bit would be a 1-1-1 (standard steel bit).

c. A bit for medium hard could be: 5-4-3. This is a roller bit for medium hard formation with gauge protected standard bearings.

The use of inappropriate bits results in early wear, poor ROP, over torque (and possible drill string failure).

]

4 Briefly describe the wear reporting notations for tooth wear, bearing wear and gauge wear of

bits. An 8.5-in Class 1-3-4 (sealed journal bearings) bit drilled over 16 hours with an average bit

weight and rotary speed of 45,000 lbf and 90 rpm, respectively. When the bit was pulled, it

was graded T-5, B-6, G-I. Compute the average formation abrasiveness and bearing constant

for the depth interval drilled. Also compute the overall bit life and the optimum bit life using

the following cost and footage data: Bit cost: $45000.00; Rig Cost: $54000/hr; Trip time: 10

hrs. [3 + 4 + 3 MARKS]

Footage, ∆∆∆∆D (ft) Drilling Time, td + tc(hrs) Remarks

0 0 New bit 30 2

50 4

65 6

77 8

85 10

90 12

95 14

100 16

102 18

104 20

105 22 Torque increased

Answer:

Given db = 8.5

Class of bit = 1-3-4

Drill depth = 100 ft with time tb = 16 hrs

Average bit weight Wb = 45,000 lbf = 45kp

Pulled out bit was graded as T-5, B-6, G-1

H1 = 1.84, H2= 6, (Wb/db)m = 8.0 kP/in

J2 = 0.0802

The average formation hardness ΤH = 16 / 0.0802[5/8+6(62/2)] = 232 hours

B1 = 1.60 and B2 = 1.0

J3 = (60/90)^1.6 x (4 x 8.5/45)^1.0 = 0.523 x 0.756 = 0.39

The bearing constant Ta = 16/ 0.39*6/8 = 30.7 hours

Bit life (Tbl)= 0.0802 x 232 x (1 + 6 x (6^2/2)) = 2028 hours.

Total bearing life (Tbb) = 0.039 x 30.7 x 1 = 12.15 hours (very short! – due to high WOB).

Bit life = min(2028,12.15) =12.15 hours. The bit life is controlled by the bearing wear.

Tt = 10 hrs, Cb = $45,000, Cr = $54,000.

Table of drilling costs (Cf) is below;

We can see the minimum cost is where the footage is 77 and drill time is 8 hours – therefore the optimum bit life is 8 hours. In practice however I would recommend 10 hours as the increase in cost for 2 more hours is only $77 (i.e. $36 per hour).

5. A driller is pulling on stuck drillstring. The derrick is capable of supporting a

maximum equivalent derrick load of 500,000 lbf, the drilling line (Fast Line) has a

strength of 51,200 lbf, and the strength of the drillpipe in tension is 396,000 lbf. If eight

lines are strung between the crown block and traveling block and safety factors of 2 are

required for the derrick, drillpipe, and drilling line, how hard (pulling force) can the

driller pull trying to free the stuck pipe? Also how much pulling force should be applied

on the fast line?

Given Fde = 500,000, n = 8 calculate W from:

WEn

EnEFd

++=

1 and W

n

nFde

+=

4

500,000 = (8+4/8)W therefore W = 500,000/(12/8) = 333,333 lbf max working load (W).

Total actual derrick load at W will be (1+0.84 + 0.841 x 8)/(0.841 x 8) x 333,333 = 424543lbf.

Given:

En

WF f =

Ff = 333,333/(0.841 x 8). Ff = 49544 lbf – i.e. this exceeds limit of drilling line (i.e. 51,200 lbf)! Therefore calculate max load on travelling block using the given fast line strength (divided by two for the safety factor):

W = Ff x En = 51200/2 x (0.841x8) = 172,236 lbf max pull before fast line strength will be exceeded.

Drillpipe pull = 396,000 / 2 i.e. 198,000 lbf max pull permissible – this is more than possible given above.

The driller should not exceed the drillpipe tension limit at any time (with safety factor) – i.e. 198,000 lbf.

Given:

En

WF f = → Ff = 198,000/(0.841 x 8)

W = 29,429

No more than 29,429 lbf force should be applied to fast line – any more exceeds the drill pipe strength.

6. Your company has premium drill pipes of E and G and class 2 drill pipes of S

grades with outside diameter 5” and inside diameter 4.27” (DP 5”x4.276”). You will be

using a 500 ft drill collar with outside diameter 7” and inside diameter 2.5” (DC 7”x 2

½”). Assume the drilling will be performed using 15 ppg mud weight. Design a

drillstring maintaining API specified minimum MOP of 100,000 for each grade to drill

at least 20,000 ft.

Grade E will be used adjacent to drill collar. We have to find out the maximum drillstring

length that comprises grade E drill pipes satisfying MOP>=100,000 lbs. Let us assume, this

length is LDPE .

Note: Weight per ft (lb/ft) data is not given for our DC and DP. So the total weight has to be

calculated from X-section, length and specific weight.

MOP = Margin of Over Pull = MAAP – HL (hook load)

MAAL (Maximum Allowable Axial Load) = (X-Area)x(Tensile Yield Strength)x(1-Wear

rate)

HL =- weight of drill collars + weight of drill pipes = HL = WDC + WDP

Knowing:

– Specific weight of steel: 490 lb/ft3, 65.5 ppg; 1ft3 = 7.481 gal

– 1- MW/65.5: buoyancy factor

LDC

LDPE

Tool joint

( )

( )

−××××−+

−××××−=

5.651490

144

1

4

5.651490

144

1

4

22

22

MWLDD

MWLDDHL

DPEIDPEODPE

DCIDCODC

π

π

( )

( )

DPE

DPE

L

L

84.1344043

5.65

151490

144

1276.45

4

5.65

151490500

144

15.27

4

22

22

+=

−××××−+

−××××−=

π

π ( )

lb8.475,316

)2.01(75000276.454

22

=

−××−=π

MAAL

Our design condition is to maintain MOP < 100,000lb

The 20,000 ft drill string must be completed with Grade G pipe.

E section total weight is 12461 ft Grade E + 500 ft drill collar.

Total weight = WDC + WDP = 44043 + (13.84 x 12461)

= 216500 lb.

LDC + LDPE + LDPG = 22,107 ft.

This is adequate for a 20,000 ft well.

LDPG = 20,000 – (500+12,461) = 7,039ft

Total weight = WDC + WDPE + WDPG

= 44,043+ 216,500 + (7,039 x 13.84) = 357,964 lb

If your designed drillstring gets stuck while drilling at 15,000 ft, how much maximum

over pull you should apply in an attempt to free the drillstring? How much over pull

could you apply if the string were stuck while drilling at 5000 ft?

ft12461

84.13

44043100000316476

100000)84.1344043(316476

100000

=

−−=

=+−

=−

∴−=

DPE

DPE

L

L

HLMAAL

HLMAALMOP

500 ft

12,461 ft

LDPG

Tool joint

Grade

change

Drill collar

Grade E

( )

DPG

DPG

EG

L

L

HL

84.13216500

5.65

151490

144

1276.45

4216500 22

+=

−×××

×−+=+

π

( )

lb1.066,443

)2.01(105000276.454

22

=

−××−=π

GMAAL

ft2.9146

84.13

216500100000443066

100000)84.13216500(443066

100000

=

−−=

=+−

=−

∴−=

+

DPG

DGP

DPG

EGG

L

L

L

HLMAAL

HLMAALMOP

LDC = 500 ft

LDPE = 12,461 ft

LDPG = 9146 ft

LDPS

Tool joint

Grade

change

MOP at 15,000 (Grade G drill pipe section is1,039 ft long i.e. 15,000-500-13,461)

MOP -= MAAL – HL

MAAL at A-A = (X-area at A-A)(Yield Strength of Grade S drillpipe)(1-Wear Rate% at A-A)

HL at A-A = Weight of LDC + weight of LDPE + weight of LDPG + weight of LDPS

MAAL @ 15,000 ft = MAALg

HL @ 15,000ft = WDC + WDPE + WDPG

= 44,043 + 216,500 + (1039 x 13.84) = 261,982 lb.

Maximum over pull @ 15,000 ft = 443,066 – 261,982 = 181,084 lb.

MOP at 5,000 (Grade E drill pipe section is4,500 ft long i.e. 5,000-500)

MOP -= MAAL – HL

MAAL at A-A = (X-area at A-A)(Yield Strength of Grade S drillpipe)(1-Wear Rate% at A-A)

HL at A-A = Weight of LDC + weight of LDPE + weight of LDPG + weight of LDPS

MAAL @ 5,000 ft = MAALE

HL @ 5,000ft = WDC + WDPE

= 44,043 + (4500 x 13.84) = 106,323 lb.

Maximum over pull @ 5,000 ft = 316,475 – 106,323 = 210,152 lb.

( )

lb066,443

)2.01(105000276.454

22

=

−××−=π

GMAAL

( )

lb475,316

)2.01(75000276.454

22

=

−××−=π

EMAAL

7. A drillstring is made up of drillpipes of a special grade for slimhole drilling with the following strength particulars:

Outside diameter, Do = 3.65 in, Inside diameter, din = 3.00 in,

Young’s modulus, E = 30×106 psi

Yield strength, σy = 110,000 psi

Shear modulus, G = 12×106 psi

Ultimate tensile strength, σu = 125,000 psi Weight per linear foot, W = 12 lb/ft.

The string has already completed drilling Event_1 with the following particulars using 10

ppg mud while passing through the following dogleg profile: The next well is planned to drill with the following particulars: Assuming that the event_2 drilling will be performed using a drilling mud of 11.30 ppg,

perform the following calculations for the pipe section, A:

1. The fatigue life consumed in event_1.

2. Is the remaining fatigue life sufficient to complete event_2? If so, how much life

will remain after the completion of event_2?

(Note: Use the S-N curve approximately that we used in Drilling System Design Lecture and use 30 ft for a drill pipe length).

Event_2 Depth to be drilled = 13,500 ft Length below the pipe section, A = 10,000 ft V = 40 ft/hr R = 150 rpm

d = 50 ft

Dogleg Profile: Station Inclination (deg.) 0.0 0.0 1 4.0 2 8.5 3 10.5

Event 1 Depth drilled = 15,000 ft Length below pipe section, A = 12,500 ft V = 20 ft/hr R = 100 rpm d = 50 ft

Dogleg profile: Station Inclination (deg.) 0 0.0 1 1.5 2 4.5 3 6.5

See working on next page.

1. Fatigue live consumed in event 1 is ~10% 2. Dmn = 0.34. This can be considered safe against fatigue. Following event 2 only

0.64 fatigue life remains.

Given: 3.00E+07 PSI

3.65 in

Outside diameter, Do = 3.65 in, 3.65

Inside diameter, din = 3.00 in, 3

Young’s modulus, E = 30×106 psi

Yield strength, σy = 110,000 psi

Shear modulus, G = 12×106 psi

Ultimate tensile strength, σu = 125,000 psi

Calculate the fatigue in Event 1:

Teff= (12,500x12) x (1 - 11.3/65.5) = 124122.1 lb Weight per linear foot, W = 12 lb/ft.

Ap = π/4(Do2 - Di2) = 3.394884 SqIn

σa = Teff/Ap= 36561.53 PSI

L = 30ft = 180 inches

d=50ft

I = 15000 ft ^ 4 = 1.05E+21 in^4

R = 100RPM

V=20ft/hr

ni = 60RD/V

Station Inclin (deg) DLS C (rad/in) σb = σad = ni Nfi Di

0 0 0 0 0.00E+00 0 15000 0 0

1 1.5 3 4.63E-05 2.53E+03 14331.57 15000 0.00E+00 0

2 4.5 6 8.72E-05 4.77E+03 26991.64 15000 1.50E+05 0.1

3 6.5 4 5.81E-05 3.18E+03 17984.11 15000 1.00E+07 0.0015

Summed Fatigue Dm = 0.1015

Calculate the fatigue in Event 2 given:

L=10,000ft

d=50ft

I=13,500ft

V=40ft/hr

R=150RPM

Teff= (10,000x12) x (1 - 11.3/65.5) = 99297.71

Ap = π/4(Do2 - Di2) = 3.394884

σa = Teff/Ap= 29249.22

Station Inclin (deg) DLS C (rad/in) σb = σad = ni Nfi Di

0 0 0 0 0.00E+00 0 11250 0 0

1 4 8 1.64E-05 8.98E+02 5076.41 11250 0.00E+00 0

2 8.5 9 1.30E-04 7.12E+03 40239.84 11250 5.00E+04 0.225

3 10.5 4 5.81E-05 3.18E+03 17984.11 11250 1.00E+06 0.01125

Summed Fatigue Dm = 0.23625

Total Fatigue:

Dm Event 1 + Dm Event 2 = 0.33775 (34% chance of failure - still safe)

E – Youngs modulus (psi)

Do – outer diameter (in)

C – dogleg severity (rad/in)

I – moment of inertia (in4)

L – half-length of one drillpipe between joints (in)

Teff – in lbau

ubad

σσ

σσσ

−×= psi,

tanh2

formula,sLubinski'From

=

EI

TL

EI

TCLED

eff

eff

o

bσ

densitySteel

densityMud1FactorBuoyancy

drillpipeofareaSectionalX

factorbuoyancydrillpipeofweightHanging

−=

−

×==

s

eff

aA

Tσ

8. The following pore pressure and fracture gradient data against drilling depth are available

from a well previously drilled in a region. Develop a casing-bit-mud program for a new well

to be drilled in the region. The target drilling depth is 15500 ft and it is decided that 5-in

production casing will be used in the well. Determine the number of casing strings, casing

setting depths, casing sizes, bit sizes and mud weights to drill to the target depth. Note

that the government legislation requires setting a casing string around 2000 ft for

environmental reasons.

Depth (ft) Pore Pressure (ppg) Fracture Gradient (ppg)

0 8.90 11.67

500 8.90 11.84

1000 8.90 12.08

1500 8.90 12.49

2000 8.90 12.85

2500 8.90 13.23

3000 8.90 13.59

3500 8.90 13.91

4000 8.90 14.25

4500 8.90 14.50

5000 8.90 14.76

5500 8.90 15.01

6000 8.90 15.23

6500 8.90 15.42

7000 8.90 15.62

7500 8.90 15.79

8000 8.90 15.98

8500 8.90 16.16

9000 8.90 16.32

9500 8.94 16.49

10000 9.12 16.73

10500 9.66 16.98

11000 11.11 17.25

11500 13.24 17.67

12000 15.07 17.97

12500 15.87 18.22

13000 16.07 18.30

13500 16.24 18.36

14000 16.34 18.44

14500 16.46 18.52

15000 16.57 18.56

15500 16.68 18.63

16000 16.78 18.69

16500 16.92 18.76

Answer:

A drilling program in essence consists of;

• Mud Program

• Bit Program

• Casing Program

• Cement Program

Mud Program: The purpose of mud is to; - Transport the cuttings - Avoid stuck pipe - Protect the reservoir - Stabilise the hole - Minimise torque - Suspend the cuttings - Minimise drag - Clean & Cool the Drill Bit - Assist Drilling But most importantly - Control pressure. Step 1: Add a trip margin (0.5 ppg) and subtract a kick margin (0.5 ppg). Step 2: Calculate W1, W2 and W3 based on minimising mud types.

From Figure 2 - Graph of PP / FG vs depth W1, W2 and W3 are determined: W1= 9.5ppg, W2 = 15ppg, W3 = 17.5 ppg Step 3: Estimate the mud volumes required to drill each constant weight section. Assuming we require 500 BBL for W1 with a cP of 15 using water, barite and bentonite.

From Figure 1 - Mud Viscosity / Weight Curve: 1 bbl of 15-cp viscosity mud will require 20 lb of bentonite. For 500 bbls of mud, total bentonite required = 500 bbl x 20 lb/bbl = 10,000 lbs Volume of bentonite = 10,000 lb/875 lb/bbl = 11.43 bbl Volume of fresh water = 500 – 11.43 = 488.6 bbl. Weight of fresh water = 488.6 bbl x 350 lb/bbl = 171,000 lb Weight of bentonite = 10,000 lb Total weight of bentonite-water slurry = 171,000+10,000 = 181,000 lb Mud weight in ppg = total weight/total gallons = 181,000/(500x42) = 8.62 ppg Given X = bbl of barite to increase weight of 100 bbl of mud to W1 (9.5 + 10% = ~10ppg) :

2

12

5.351490

W

WWX

−

−=

lb/bbl5.197

bbl 100per sacks5.1970.105.35

62.80.101490

=

=−

−=X

So, for 500 bbls of mud, we need 500x197.Volume increase due to barite = 987.49Total mud volume we hav

Calculate the following to prepare approximately 500 barrels of W1 lb/gal drilling mud with 15 cp

viscosity:

1. Amount of fresh water

2. Amount of clay (bentonite)

3. Amount of barite –

Calculate the amounts of additional barite required to increase the mud weights from W1 to W2

and then from W2 to W3 per 100 bbls of mud to drill the next two sections.

Knowing W1= 9.5ppg, W2 = 15p

Figure 1 - Mud Viscosity / Weight Curve

lb/bbl8.399

sacks8.399155.35

5.915149021

=

=−

−=→X

lb/bbl9.206

sacks9.2065.175.35

155.17149032

=

=−

−=→X

0 bbls of mud, we need 500x197.5 = 98749 lbs or 987.49 sacks of barite.increase due to barite = 987.49 sacks / 14.9 sacks/bbl = 66.27 bbls of barite.

Total mud volume we have now 500+66.27 = 566.27 bbls

Calculate the following to prepare approximately 500 barrels of W1 lb/gal drilling mud with 15 cp

Amount of fresh water – 488.6lbs

Amount of clay (bentonite) - 10,000lbs

– 987.49 sacks

Calculate the amounts of additional barite required to increase the mud weights from W1 to W2

and then from W2 to W3 per 100 bbls of mud to drill the next two sections.

W1= 9.5ppg, W2 = 15ppg, W3 = 17.5 ppg

bbl 100per sacks

bbl 100per sacks

sacks of barite. bbls of barite.

Calculate the following to prepare approximately 500 barrels of W1 lb/gal drilling mud with 15 cp

Calculate the amounts of additional barite required to increase the mud weights from W1 to W2

Figure 2 - Graph of PP / FG vs depth

Intent of casing design:

Consolidate: that is to ensure hole stability to casing shoe Separate and/or protect: for example to seperate multiple pressure or production zones Pressure control: ensure strength in the hole above For production: to produce through Pressure Gradients plotting is a way to determine casing setting depths and mud weights versus depth. Using Figure 2 - Graph of PP / FG vs depth the following casing design has been determined (taking into constraint the government regulation to have a 2000ft casing conductor).

Figure 3 - Initial Casing Design

Casing Sizes and Bit Size calculations (from tables).

Production casing 5 in. (specified by production engineer); coupling size = 5.563-in

Bit size to drill a hole in which this casing can be run, 6 1/2-in (Table 7.7)

This 6 1/2-in bit has to pass through the intermediate casing

The intermediate casing size through which the above bit would pass, 7 5/8-in (Table 7.8)

Bit size to drill a hole in which the 7 5/8-in casing can be run, 9 7/8 -in (Table 7.7).

This 9 7/8-in bit has to pass through the surface casing

The surface casing size through which the above bit would pass, 10 3/4-in (Table 7.8)

The bit size to drill a hole in which the 10 3/4-in casing can be run, 15 -in (Table 7.7)

This 15-in bit has to pass through the conductor pipe

The conductor pipe size through which the above bit would pass, 16 -in (Table 7.8)

The bit size to drill a hole in which the 16 -in conductor pipe can be run, 20-in.

Design summary for Casing and Bit:

Depth

(ft

Mud Density

(ppg)

Casing Type Casing Size

(in)

Bit Size

(in)

180 ~ 9 Conductor 16 20

6000 ~ 9.5 Surface 10 3/4 15

12000 15 Intermediate 7 5/8 9 7/8

15500 17.5 Production 5 6 ½

Optimized casing design:

The above casing design meets the environmental specifications (i.e. for surface casing to be set to

2000 ft rather than 6000 ft) but has greatly reduced production tubing length.

Design the 2nd

deepest casing string against burst, collapse and tensile loads with safety factors of

1.3, 1.0 and 1.8, respectively. Assume a 15 ppg cement column of 1000 ft height at the casing shoe.

Note: 2nd

deepest casing is the Intermediate 7 5/8” casing.

Burst Pressure Calculation

PI – PO = Burst Pressure

PI = Reservoir Pressure – weight of gas column

To calculate Reservoir Pressure: pressure in psi = ppg x 0.052 x True Vertical Depth (TVD) in ft =

16.7 x 0.052 x 15500 = 13,460 PSI

PI = 13,460 – 15,500 x 0.1 = 11,910 psi (at surface)

PI = 13,460 – (12,000-1000) x 0.1 = 12,360 psi (at top of cement)

PI = 13,460 – (12,000 x 0.1) = 12,260 psi (at casing shoe)

PO = weight of fluid in annulus (assume mix water of ppg 8.5)

PO = 0 PSI (at surface)

PO = 0.052 x 11000 x 8.5 = 4862 PSI (at top of cement)

PO = 4862 + (0.052 x 1000 x 8.5) = 8.5) = 5304 PSI (at casing shoe)

PI – PO = 11,910 – 0 = 11,910 PSI (at surface)

PI – PO = 12,360 – 4862 = 7,498 PSI (at top of cement)

PI – PO = 12,260 – 5,304 = 6,956 PSI (at casing shoe)

Maximum burst pressure is 11,910 – with safety factor of 1.3 this is 15,483 PSI.

Collapse Pressure Calculation

PO – PI = Collapse Pressure

PI = 0 (i.e. evacuated casing at atmospheric pressure).

PO = weight of fluid in annulus

= 0.052 x (12000 – 1000) x 10 ppg = 5720 psi (at top of cement)

= 5720 + 0.052 x 1000ft x 15ppg = 6500 psi (at casing shoe).

Collapse pressure = 5720 – 0 = 5720 psi (at top of cement) and 6500 – 0 = 6500 psi (at casing shoe).

Maximum collapse pressure is 6,500 PSI – with a safety factor of 1.0 this 6,500 PSI.

Figure 4 - 7 5/8 Casing Table

Burst pressure of 15,483 PSI is our first limiting factor. We need to use grade P-110 with 0.625 wall

thickness minimum (with a collapse pressure of 16,550 PSI and burst pressure of 15,780).

Tensile strength calculation – the selected grade has a yield strength of 1,512 klbf.

The self weight of casing hanging in the hole:= 15,500 x 47.1 #/ft = 730,050 lbs

- buoyancy effect assume = 0

+ bending load assume = 0 (straight hole)

+ shock load (at running speed in ft/s)

+ pressure loads

+ overpull loads (when static) = 100,000 lbs

Total axial load ~ 830 klbs

Total axial load x design safety factor= 830 x 1.8 = 1089 klbs - we have a margin of 440 klbs

– tensile load is safe.